A STEM literate person considers how STEM can improve the social, cultural, economic, and environmental conditions of their local and global communities.

(b) Galileo: Inquiry-based learning

Galileo.org is mostly aimed at teachers, and provides some interesting inquiry-based resources in mathematics and science, including lesson designs and classroom examples. There are also articles and suggestions regarding technology-based approaches to teaching.

3. Math in the news

Consider a 3,4,5 Pythagorean triple, as follows. We color 2 of the squares red, and the other blue.

Next, construct a 5,12,13 right triangle based on that blue 5×5 square, as follows.

We choose blue and red for the colors of the other two triangles. We could have swapped those 2 colors, but we can't have all blue, or all red.

What we're doing is demonstrating Ronald Graham's Boolean Pythagorean Triples problem, which asks if it's possible to color the squares given by all the Pythagorean triples using exactly two colors only.

With small numbers like the ones above, it seems quite straightforward, but as the numbers get bigger we may be forced to choose a color for one of the triples based on a previous condition, only to find the current set doesn't allow that color (because we may end up with all blue, or all red squares).

Graham offered a $100 prize to anyone who could solve the problem (that is, can it be done for all Pythagorean triples?).

Last month, researchers Marijn Heule from the University of Texas, Victor Marek from the University of Kentucky, and Oliver Kullmann from Swansea University used a Stampede supercomputer to solve it.

The answer is a 200 terabyte file which is not human-readable. It would take you 30,000 hours just to download it!

So is it a proof?

This is a great example of how modern mathematics is becoming irrevocably entwined with computers. There has been some interesting discussion about whether such a large file can be considered a "proof" at all. It can be read by computers and verified by computers, but we humans can't really step through such a proof and check its validity.

4. Math puzzles

Correct answers with sufficient explanation (which included some reasoning about the number of solutions) were given by: Don, Giorgos, Lidia and Nour. Other partially correct answers (not including the uniqueness part) were given by Cara and Manju.

Nour's approach was to graph it (after substracting the 3 right-hand terms from both sides). He would have seen something like this:

Graphical solution of equation

The graph is discontinuous, since it has no value for x < 1988. It continues on beyond 2100 getting less and less steep, but never crosses the x-axis again. (It looks something like a half-parabola on its side.) So there was a unique solution.

New math puzzle

At a plant stall, I can buy 8 xerophytes for $1, yams for $1 each and zinnias for $10 each. I grab a selection of plants and find at the checkout that I have exactly 100 plants and they will cost me $100. How many of each type did I get?

From the cost of the plants, we get the following equation:
and from the total number of plants, we have:

Solving the second equation for and substituting into the first (and multiplying the whole thing by 8), we get:

This equation has infinitely many solutions in and , but we are only interested in integer solutions (can't buy partial plants), making this a Diophantine Equation. More specifically, we want non-negative integer solutions, because we cannot buy negative plants. Using the Euclidean Algorithm, we can find all of the non-negative integer solutions : and .

Plugging these into either of the first 2 equations, we get solutions of: and

Assuming that "a selection of plants" means that you got some of each (each is greater than zero), then we eliminate the trivial solution of buying 100 yams, and the only possible remaining solution is:72 xerophytes
21 yams
7 zinnias

Let the number of xerophytes,yams and zinnias are x,y and z respectively.
According to the given
x+y+z=100...................A
price of xerophytes 1/8 $,so price of x xerophytes will be (1/8)x $
Similarly prices of y yams and z zinniats will be y $ and 10z $
Total price is 100$
Hence. x/8+y+10z=100.................B
A-B gives 7x/8-9z=0
7x=72z
X=72z/7....................................C
Since x is a whole number(number of plant may not be fraction)
7 divide 72 or 7 divides z
But 7 does not divide 72
So 7 must divede z i-e z is multiple of 7
z=0,7,14...
Assuming that number of plants of any kind is not equal to 0
and we can purchase at most 10 zinniats from 100$,this implies
0 < z < 10
So z=7 (The above inequality excludes all multiples except 7 itself)
Putting z=7 In C
x=72*7/7=72
Putting x=72 and y=7 in A
72+y+7=100
y=21
Hence number of xerophytes, yams and zinnias are 72, 21 and 7